Question

Two blocks of masses m₁ and m₂ are joined by a wire of Young’s modulus Y via a massless pulley. The area of cross-section of the wire is S and its length is L. When the system is released, increase in length of the wire is

• A. $\frac{m_{1}m_{2}gL}{YS(m_{1} + m_{2})}$
• B. $\frac{2m_{1}m_{2}gL}{YS(m_{1} + m_{2})}$
• C. $\frac{(m_{1} - m_{2})gL}{YS(m_{1} + m_{2})}$
• D. $\frac{4m_{1}m_{2}gL}{YS(m_{1} + m_{2})}$

Answer 1

Answer: (B)

$\frac{2m_{1}m_{2}gL}{YS(m_{1} + m_{2})}$

Answer 2

Tension in the wire $T = \frac{2m_{1}m_{2}}{m_{1} + m_{2}}g$

$\therefore$stress in the wire $= \frac{T}{S} = \frac{2m_{1}m_{2}g}{S(m_{1} + m_{2})}$

∴ Strain $\frac{l}{L} = \frac{\text{Stres}s}{Y} = \frac{2m_{1}m_{2}g}{YS(m_{1} + m_{2})}$ ⇒ $l = \frac{2m_{1}m_{2}gL}{YS(m_{1} + m_{2})}$